Find $\lim_{x\to\infty}\dfrac{x-3}{\cos(x)+x}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-3$ (Choice B) B $1$ (Choice C) C $0$ (Choice D) D The limit doesn't exist
When dealing with limits that include $\cos(x)$, it's important to remember that $\lim_{x\to\infty}\cos(x)$ doesn't exist, as $\cos(x)$ keeps oscillating between $-1$ and $1$ forever. ${2}$ ${4}$ ${6}$ ${8}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ ${2}$ $y$ $x$ $y=\cos(x)$ This doesn't necessarily mean that our limit doesn't exist. Think what happens to $\dfrac{x-3}{\cos(x)+x}$ as $x$ increases towards positive infinity. $\cos(x)$ oscillates between $-1$ and $1$. This can be represented mathematically by the following double inequality: $\dfrac{x-3}{-1+x}\leq\dfrac{x-3}{\cos(x)+x}\leq\dfrac{x-3}{1+x}$ The result is a graph that's always between the graphs of $y=\dfrac{x-3}{\pm 1+x}$ (the dashed lines). ${5}$ ${10}$ ${\llap{-}5}$ ${\llap{-}10}$ ${5}$ ${10}$ ${\llap{-}5}$ ${\llap{-}10}$ $y$ $x$ Since $\lim_{x\to\infty}\dfrac{x-3}{\pm 1+x}=1$, so must our limit be equal to $1$. In conclusion, $\lim_{x\to\infty}\dfrac{x-3}{\cos(x)+x}=1$.